Finding Points of Intersection of Two Lines

Suppose that we have two lines. If these two lines intersect, then sometimes it might be important to find the coordinates of this intersection. For example, consider the lines $y = 2x$ and $y = -x + 6$:

We can clearly see that an intersection occurs at the point $(2, 4)$. Unfortunately, it is not always practical to graph the lines in order to determine the coordinates of their intersection. We will now look at some methods for calculating the coordinates of an intersection:

Method 1: Substitution

Suppose we have two lines in the form $y = mx + b$. For example, the lines $L_1: y = 2x$ and $L_2: y = -x + 6$. First take any of the lines. We will take $L_1$. We note that from $L_2$, that $y = -x + 6$. If we make this substitution into $L_1$, we get that:

(1)

\begin{align} y = 2x \\ -x + 6 = 2x \end{align}

Now all we have to do is solve this 1-variable linear equation and we will get the x-coordinate of our intersection:

(2)

\begin{align} -x + 6 = 2x \\ 6 = 3x \\ 2 = x \end{align}

Thus the x-coordinate of our intersection is 2 (which we verified earlier). To find the x-coordinate, we will now take any of the lines and set $x$ to be 2 to get the corresponding $y$ coordinate. We note that the choice of the equation doesn't matter, though it is usually best to pick the easier equation. Let's choose $L_1$ again:

(3)

\begin{align} y = 2x \\ y = 2(2) \\ y = 4 \end{align}

Therefore, the coordinates to which our lines intersect is $(2, 4)$.

Method 2: Addition/Subtraction

If we have two lines in the form $ax + by + c = 0$, we can do some multiplication and addition/subtraction between the two equations to isolate for a variable, solve for it, and then plug it back into either equation again to get the corresponding y-coordinate. Once again, let's look at the equations $L_1: 2x - y + 0 = 0$ and $L_2: x + y - 6 = 0$ (the same equations as before.

If we take $L_1$ and add $L_2$, we end up eliminating the $y$-variable:

Now solving this equation we get that $x = -\frac{4}{7}$, and substituting this value in to the first equation, we get that $y = -\frac{5}{7}$. Therefore the coordinates of our intersection are $(-\frac{4}{7}, -\frac{5}{7})$

3. Since both lines are in $y = mx + b$ form, setting them equal we get that:

(9)

\begin{align} mx + 1 = mx + 2 \\ 1 = 2 \end{align}

This is an inconsistency though, and hence there is no point of intersection.